Acceleration in the Electric Field Calculator

See how fast a charged particle accelerates in a uniform electric field, and how much speed it gains crossing a field or a pair of charged plates.

⚡ Acceleration in the Electric Field Calculator
Particle
Charge (q)
e
Mass (m)
u
Electric field (E)
V/m
Distance travelled (s)
m
Plate voltage (V)
V
Plate gap (d)
mm
Acceleration a = qE/m
—
Force |q| E
—
Field used
—
Voltage crossed
—
Energy gained
—
Force direction
—
Speed, non-relativistic
—
Speed, relativistic
—
Speed as a fraction of c
—
Lorentz factor
—
Time to cross, classical
—
Time to cross, relativistic
—

⚡ What is Acceleration in an Electric Field?

Acceleration in an electric field is the change of velocity that a charged particle undergoes because the field pushes on it. An electric field E exerts a force F = qE on a charge q. Newton's second law then gives the acceleration a = F/m = qE/m. A positive charge accelerates in the direction of the field. A negative charge, such as an electron, accelerates the opposite way. In a uniform field the acceleration is constant, so the motion from rest is the same as a falling object under steady gravity, only the pull depends on the charge to mass ratio.

This idea sits behind a surprising number of devices. Old cathode ray tubes and modern electron microscopes use an electron gun that accelerates electrons across thousands of volts. Mass spectrometers and ion thrusters push ions with electric fields, linear accelerators use sequences of gaps, and X-ray tubes drive electrons into a metal target. The same physics describes ink droplets deflected in a printer, charged dust in a plasma, and the motion of ions in a drift tube.

Some points are worth keeping straight. The acceleration depends on q/m, not on q or m alone, so a heavier particle can be accelerated just as strongly if it carries enough charge. The energy gained depends only on the voltage crossed, W = |q|V, so two plate sets with different gaps but the same voltage give the same final energy. The classical speed formula fails when the speed gets within a few tenths of the speed of light, because mass-energy grows and the speed can never reach c. That is why this tool shows a relativistic result next to the classical one.

The calculator offers two modes. In the first you enter a field strength and the distance travelled. In the second you enter a plate voltage and gap, and it finds E = V/d. Both report the acceleration, force, energy in electron volts, crossing time and final speed, and the chart shows how the speed rises towards the limit of c.

📐 Formula

a  =  q E ÷ m
a = acceleration (m/s²), along E for q > 0 and against E for q < 0
q = charge (C). The calculator takes it in units of e, with e = 1.602176634×10−19 C
E = electric field (V/m). For two plates, E = V ÷ d
m = mass (kg). The calculator takes it in atomic mass units, with 1 u = 1.66053906660×10−27 kg
Energy gained: K = |q| E s = |q| V, and the classical speed is v = √(2 K / m) = √(2 a s)
Relativistic: γ = 1 + K / (m c²), v = c √(1 − 1/γ²), and t = (c / a) √(γ² − 1)
Example: electron, E = 1000 V/m: a = 1.602×10−19 × 1000 ÷ 9.109×10−31 = 1.7588×1014 m/s²

📖 How to Use This Calculator

Steps

1
Choose how the field is set. Pick Field and distance if you know the field strength in V/m and the distance travelled, or Plate voltage and gap if the field comes from two charged plates.
2
Pick the particle. Choose an electron, proton or alpha particle, or select Custom and type a charge in units of e and a mass in atomic mass units.
3
Read the acceleration, speed and chart. Read the acceleration, force, energy gained, crossing time and final speed, and check the warning if the classical speed is above 0.1 c.

💡 Example Calculations

Example 1 — Electron in a 1000 V/m Field

An electron starts from rest and travels 0.1 m in a field of 1000 V/m

1
F = e E = 1.602×10−19 × 1000 = 1.6022e-16 N
2
a = F / m = 1.6022×10−16 ÷ 9.109×10−31 = 1.7588e+14 m/s², directed against the field
3
It gains 1000 × 0.1 = 100 eV, so v = √(2 a s) = 5.9310e+6 m/s (0.0198 c) in 3.3721e-8 s. The relativistic speed is 5.9301e+6 m/s.
Acceleration = 1.7588e+14 m/s², speed = 5.9310e+6 m/s
Try this example →

Example 2 — Proton Between Plates

A proton crosses plates 5 mm apart with 10000 V between them

1
E = V / d = 10000 / 0.005 = 2.0000e+6 V/m
2
a = e E / m_p = 3.2044×10−13 ÷ 1.6726×10−27 = 1.9158e+14 m/s²
3
K = e V = 10 keV, so v = √(2 K / m) = 1.3841e+6 m/s (0.0046 c), reached in 7.2248e-9 s
Final speed = 1.3841e+6 m/s after crossing 10 kV
Try this example →

Example 3 — Fast Electron Where Relativity Matters

An electron crosses a gap of 20 mm at 100 kV, as in an X-ray tube

1
E = 100000 / 0.02 = 5.0000e+6 V/m and a = e E / m = 8.7941e+17 m/s²
2
Classical: v = √(2 e V / m) = 1.8755e+8 m/s, which is 0.6256 c, well above the 0.1 c guideline
3
Relativistic: γ = 1 + 100 keV / 510.999 keV = 1.195695, so v = c √(1 − 1/γ²) = 1.6435e+8 m/s = 0.5482 c
Correct speed = 0.5482 c; the classical formula would give 0.6256 c
Try this example →

Example 4 — Alpha Particle in a Strong Field

An alpha particle (charge 2e, mass 4.0015 u) in 1,000,000 V/m over 0.1 m

1
F = 2 e E = 3.2044e-13 N
2
a = F / m = 3.2044×10−13 ÷ 6.6447×10−27 = 4.8225e+13 m/s²
3
K = 2 e × 105 V = 200 keV, so v = 3.1056e+6 m/s (0.0104 c) after 6.4399e-8 s
Acceleration = 4.8225e+13 m/s², speed = 3.1056e+6 m/s
Try this example →

Example 5 — A Custom Particle (Muon)

A muon (charge −1 e, mass 0.1134289256 u) in 500,000 V/m over 0.4 m

1
m = 0.1134289256 × 1.66054×10−27 = 1.8835×10−28 kg and F = e E = 8.0109e-14 N
2
a = F / m = 4.2531e+14 m/s², against the field
3
The voltage crossed is 500000 × 0.4 = 200 kV, so K = 200 keV. Classical v = 1.8446e+7 m/s (0.0615 c), relativistic v = 1.8420e+7 m/s (0.0614 c).
Acceleration = 4.2531e+14 m/s², relativistic speed = 1.8420e+7 m/s
Try this example →

❓ Frequently Asked Questions

What is the formula for acceleration in an electric field?+
The acceleration is a = qE/m, from the force F = qE and Newton's second law. Here q is the charge in coulombs, E is the field in V/m and m is the mass in kg. For an electron in a field of 1000 V/m, a = 1.7588e14 m/s².
How do you find the acceleration of an electron between two plates?+
Divide the plate voltage by the gap to get the field, E = V/d, then use a = eE/m. For 100 kV across 20 mm, E = 5 million V/m and the acceleration is 8.7941e17 m/s², which is why a relativistic check is also shown.
Why does a proton accelerate less than an electron?+
The force on each is the same size e E, but the proton is 1836 times heavier. So the proton acceleration is the electron value divided by 1836. In 2 million V/m a proton reaches 1.9158e14 m/s² and an electron reaches 3.5176e17 m/s².
How fast does a particle go after crossing a voltage?+
From rest, the speed follows from energy: v = sqrt(2 |q| V / m) when v is well below the speed of light. A proton crossing 10 kV reaches 1.3841e6 m/s. An electron crossing 100 V reaches 5.931e6 m/s.
When do I need relativity for an accelerated particle?+
When the speed passes about 0.1 c (30000 km/s), the classical formula overestimates the speed by more than about one percent. An electron reaches 0.1 c at about 2.6 kV. For 100 kV the classical answer gives 0.6256 c, while the correct value is 0.5482 c.
Does the direction of the field matter for a negative charge?+
Yes. The force on a positive charge points along E, and the force on a negative charge points against E. The calculator reports the direction, and the magnitude of the acceleration is the same either way.
How is the energy gained related to voltage?+
The kinetic energy gained is the charge times the potential difference crossed. A particle with charge 2e, like an alpha, gains 200 keV across 100 kV. Electron volts make this easy: the energy in eV equals the voltage times the charge number.
How long does it take to cross the gap?+
At constant acceleration from rest, t = sqrt(2 d / a). For a proton across 5 mm at 10 kV, t = 7.2248e-9 s. The calculator also gives the relativistic crossing time, which is slightly longer for fast particles.
Is the electric field between parallel plates uniform?+
Away from the edges, yes. A uniform field E = V/d is a good model when the gap is much smaller than the plate size. Near the edges the field fringes, and a real particle gun uses shaped electrodes to control it.
Can I use a charge or mass that is not an electron, proton or alpha?+
Yes. Choose Custom and type the charge in units of e and the mass in atomic mass units. A muon, for instance, has charge -1 and mass 0.1134289 u, and a singly charged helium ion has charge 1 and mass 4.0026 u.
Is gravity included in the acceleration?+
No. For electrons and ions the electric force is many orders of magnitude larger than weight. An electron in 1000 V/m feels a force of 1.6e-16 N, while its weight is about 9e-30 N, so gravity can be ignored in almost every case.

What is the formula for acceleration in an electric field?

The acceleration is a = qE/m, from the force F = qE and Newton's second law. Here q is the charge in coulombs, E is the field in V/m and m is the mass in kg. For an electron in a field of 1000 V/m, a = 1.7588e14 m/s².

How do you find the acceleration of an electron between two plates?

Divide the plate voltage by the gap to get the field, E = V/d, then use a = eE/m. For 100 kV across 20 mm, E = 5 million V/m and the acceleration is 8.7941e17 m/s², which is why a relativistic check is also shown.

Why does a proton accelerate less than an electron?

The force on each is the same size e E, but the proton is 1836 times heavier. So the proton acceleration is a_p = a_e / 1836. In 2 million V/m a proton reaches 1.9158e14 m/s² and an electron reaches 3.5176e17 m/s².

How fast does a particle go after crossing a voltage?

From rest, the speed follows from energy: v = sqrt(2 |q| V / m) when v is well below the speed of light. A proton crossing 10 kV reaches 1.3841e6 m/s. An electron crossing 100 V reaches 5.931e6 m/s.

When do I need relativity for an accelerated particle?

When the speed passes about 0.1 c (30000 km/s), the classical formula overestimates the speed by more than about one percent. An electron reaches 0.1 c at about 2.6 kV. For 100 kV the classical answer gives 0.6256 c, while the correct value is 0.5482 c.

Does the direction of the field matter for a negative charge?

Yes. The force on a positive charge points along E, and the force on a negative charge points against E. The calculator reports the direction, and the magnitude of the acceleration is the same either way.

How is the energy gained related to voltage?

The kinetic energy gained is the charge times the potential difference crossed. A particle with charge 2e, like an alpha, gains 200 keV across 100 kV. Electron volts make this easy: the energy in eV equals the voltage times the charge number.

How long does it take to cross the gap?

At constant acceleration from rest, t = sqrt(2 d / a). For a proton across 5 mm at 10 kV, t = 7.2248e-9 s. The calculator also gives the relativistic crossing time, which is slightly longer for fast particles.

Is the electric field between parallel plates uniform?

Away from the edges, yes. A uniform field E = V/d is a good model when the gap is much smaller than the plate size. Near the edges the field fringes, and a real particle gun uses shaped electrodes to control it.

Can I use a charge or mass that is not an electron, proton or alpha?

Yes. Choose Custom and type the charge in units of e and the mass in atomic mass units. A muon, for instance, has charge -1 and mass 0.1134289 u, and a singly charged helium ion has charge 1 and mass 4.0026 u.